Does the variance of a strictly monotonically increasing function of a random variable have anything to do with the variance of the random variable? [closed]

Question

Assume that there is a continuous random variable x, and its variance is var(x). Furthermore, there is a strictly monotonically increasing function f. Can anybody prove that the larger the var(x), the larger the variance of f(x), i.e. var(f(x))? One thing important is that the expectation of this random variable is fixed, with its variance to be the only part changeable.

Answer 1

It's not true. Take $f(x) = x^2$. Let $X$ take the values $0$ and $4$ each with probability $1/2$, and $Y$ take the values $100$ and $102$ each with probability $1/2$. Then we have $\newcommand{\Var}{\operatorname{Var}}\Var(X) = 4$ and $\Var(Y)=1$, but $\Var(f(X)) = 64$ and $\Var(f(Y)) = 40804$.

For an example going the other way, try $f(x) = x^{1/2}$.

Answer 2

The Answer is still no even if all variables have mean zero. Take $f(x)=x|x|$. Let $X$ take values $-5,-1,1,5$, equally likely, and let $Y$ take values $-4,4$, equally likely. Then $Y$ has greater variance than $X$ but this is reversed when you apply $f$ to both variables. All this remains true for continuous variables as well. Add an independent variable $U$, uniform in $[-\epsilon, \epsilon ]$, to both $X$ and $Y$.